The sum and difference rules let you find the derivative of a function by differentiating each term separately. This page focuses on combining basic derivative rules with addition and subtraction, including power rule, constant rule, trig derivatives, exponential derivatives, and logarithmic derivatives. These problems help build fluency with breaking larger expressions into smaller derivative steps.
Notes
Problems
Find the derivative of each function
\(\textbf{1)}\) \(f(x)=3x^4-2x+8\)
The answer is \(12x^3-2\)
\(\,\,\,\,\,\text{To differentiate } f(x) = 3x^4 – 2x + 8, \text{ apply the power rule to each term.}\)
\(\,\,\,\,\,f'(x) = 3 \cdot 4x^{3} – 2 + 0\)
\(\,\,\,\,\,f'(x) = 12x^3 – 2\)
\(\,\,\,\,\,\)The answer is \(12x^3-2\)
\(\textbf{2)}\) \( f(x)=5-\frac{1}{x} \)
The derivative is \( f’ (x)=\displaystyle\frac{1}{x^{2}} \)
\(\,\,\,\,\,\text{To differentiate } f(x) = 5 – \frac{1}{x}, \text{ rewrite } \frac{1}{x} \text{ as } x^{-1} \text{ and use the power rule.}\)
\(\,\,\,\,\,f(x) = 5 – x^{-1}\)
\(\,\,\,\,\,f'(x) = 0 – \left(-1x^{-2}\right)\)
\(\,\,\,\,\,f'(x) = \frac{1}{x^{2}}\)
\(\,\,\,\,\,\)The derivative is \( f’ (x)=\displaystyle\frac{1}{x^{2}} \)
\(\textbf{3)}\) \( f(x)=\frac{1}{2} x^{3}+4x^{2} \)
The derivative is \( f'(x)=\displaystyle\frac{3x^{2}}{2}+8x \)
\(\,\,\,\,\,\text{Differentiate } f(x) = \frac{1}{2} x^3 + 4x^2 \text{ by applying the power rule to each term.}\)
\(\,\,\,\,\,f'(x) = \frac{1}{2} \cdot 3x^{2} + 4 \cdot 2x\)
\(\,\,\,\,\,f'(x) = \frac{3x^{2}}{2} + 8x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\displaystyle\frac{3x^{2}}{2}+8x \)
\(\textbf{4)}\) \( f(x)=\sin x-\cos x+2 \)
The derivative is \( f'(x)=\cos x+\sin x \)
\(\,\,\,\,\,\text{Differentiate } f(x) = \sin x – \cos x + 2 \text{ by applying the derivatives of } \sin x, \cos x, \text{ and constants.}\)
\(\,\,\,\,\,f'(x) = \cos x + \sin x + 0 \)
\(\,\,\,\,\,f'(x) = \cos x + \sin x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\cos x+\sin x \)
\(\textbf{5)}\) \( f(x)=5 \sin x+3x \)
The derivative is \( f'(x)=5 \cos x+3 \)
\(\,\,\,\,\,\text{Differentiate } f(x) = 5 \sin x + 3x \text{ by applying the derivatives of } \sin x \text{ and } x. \)
\(\,\,\,\,\,f'(x) = 5 \cdot \cos x + 3 \cdot 1\)
\(\,\,\,\,\,f'(x) = 5 \cos x + 3\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=5 \cos x+3 \)
\(\textbf{6)}\) \( f(x)=4-\displaystyle\frac{1}{4x^{2}} \)
The derivative is \( f'(x)=\displaystyle\frac{1}{2x^{3} } \)
\(\,\,\,\,\,\text{Rewrite } f(x) = 4 – \frac{1}{4x^2} \text{ as } f(x) = 4 – \frac{1}{4}x^{-2} \text{ and use the power rule.}\)
\(\,\,\,\,\,f(x)=4-\frac{1}{4}x^{-2}\)
\(\,\,\,\,\,f'(x) = 0 – \left(\frac{1}{4}\cdot(-2)x^{-3}\right)\)
\(\,\,\,\,\,f'(x) = \frac{1}{2}x^{-3}\)
\(\,\,\,\,\,f'(x) = \frac{1}{2x^3}\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\displaystyle\frac{1}{2x^{3} } \)
\(\textbf{7)}\) \( f(x)=\sqrt{x}+x^3 \)
The derivative is \( f'(x)=\displaystyle\frac{1}{2\sqrt{x}}+3x^2 \)
\(\,\,\,\,\,\text{Rewrite } \sqrt{x} \text{ as } x^{1/2} \text{ and apply the power rule to each term.}\)
\(\,\,\,\,\,f(x)=x^{1/2}+x^3\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}x^{-1/2}+3x^2\)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}+3x^2\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\displaystyle\frac{1}{2\sqrt{x}}+3x^2 \)
\(\textbf{8)}\) \( f(x)=\displaystyle\frac{6}{x^2}-4\sqrt{x} \)
The derivative is \( f'(x)=-\displaystyle\frac{12}{x^3}-\frac{2}{\sqrt{x}} \)
\(\,\,\,\,\,\text{Rewrite the function using exponents, then apply the power rule.}\)
\(\,\,\,\,\,f(x)=6x^{-2}-4x^{1/2}\)
\(\,\,\,\,\,f'(x)=6(-2)x^{-3}-4\left(\frac{1}{2}\right)x^{-1/2}\)
\(\,\,\,\,\,f'(x)=-12x^{-3}-2x^{-1/2}\)
\(\,\,\,\,\,f'(x)=-\frac{12}{x^3}-\frac{2}{\sqrt{x}}\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=-\displaystyle\frac{12}{x^3}-\frac{2}{\sqrt{x}} \)
\(\textbf{9)}\) \( f(x)=e^x+\ln x \)
The derivative is \( f'(x)=e^x+\frac{1}{x} \)
\(\,\,\,\,\,\text{Use the derivative rules } \frac{d}{dx}(e^x)=e^x \text{ and } \frac{d}{dx}(\ln x)=\frac{1}{x}. \)
\(\,\,\,\,\,f'(x)=e^x+\frac{1}{x}\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=e^x+\frac{1}{x} \)
\(\textbf{10)}\) \( f(x)=\tan x-4\cos x \)
The derivative is \( f'(x)=\sec^2 x+4\sin x \)
\(\,\,\,\,\,\text{Use } \frac{d}{dx}(\tan x)=\sec^2 x \text{ and } \frac{d}{dx}(\cos x)=-\sin x. \)
\(\,\,\,\,\,f'(x)=\sec^2 x-4(-\sin x)\)
\(\,\,\,\,\,f'(x)=\sec^2 x+4\sin x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\sec^2 x+4\sin x \)
\(\textbf{11)}\) \( f(x)=x^5-\displaystyle\frac{3}{x}+\cos x \)
The derivative is \( f'(x)=5x^4+\frac{3}{x^2}-\sin x \)
\(\,\,\,\,\,\text{Rewrite } -\frac{3}{x} \text{ as } -3x^{-1}, \text{ then differentiate each term.}\)
\(\,\,\,\,\,f(x)=x^5-3x^{-1}+\cos x\)
\(\,\,\,\,\,f'(x)=5x^4-3(-1)x^{-2}-\sin x\)
\(\,\,\,\,\,f'(x)=5x^4+3x^{-2}-\sin x\)
\(\,\,\,\,\,f'(x)=5x^4+\frac{3}{x^2}-\sin x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=5x^4+\frac{3}{x^2}-\sin x \)
\(\textbf{12)}\) \( f(x)=7x^6-4x^3+x \)
The derivative is \( f'(x)=42x^5-12x^2+1 \)
\(\,\,\,\,\,\text{Differentiate each term using the power rule.}\)
\(\,\,\,\,\,f'(x)=7\cdot6x^5-4\cdot3x^2+1\)
\(\,\,\,\,\,f'(x)=42x^5-12x^2+1\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=42x^5-12x^2+1 \)
\(\textbf{13)}\) \( f(x)=\displaystyle\frac{2}{x^3}+5x^2-9 \)
The derivative is \( f'(x)=-\displaystyle\frac{6}{x^4}+10x \)
\(\,\,\,\,\,\text{Rewrite } \frac{2}{x^3} \text{ as } 2x^{-3}. \)
\(\,\,\,\,\,f(x)=2x^{-3}+5x^2-9\)
\(\,\,\,\,\,f'(x)=2(-3)x^{-4}+10x+0\)
\(\,\,\,\,\,f'(x)=-6x^{-4}+10x\)
\(\,\,\,\,\,f'(x)=-\frac{6}{x^4}+10x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=-\displaystyle\frac{6}{x^4}+10x \)
\(\textbf{14)}\) \( f(x)=\sqrt{x}-\displaystyle\frac{5}{x}+2x^4 \)
The derivative is \( f'(x)=\displaystyle\frac{1}{2\sqrt{x}}+\frac{5}{x^2}+8x^3 \)
\(\,\,\,\,\,\text{Rewrite radicals and fractions using exponents.}\)
\(\,\,\,\,\,f(x)=x^{1/2}-5x^{-1}+2x^4\)
\(\,\,\,\,\,f'(x)=\frac{1}{2}x^{-1/2}-5(-1)x^{-2}+8x^3\)
\(\,\,\,\,\,f'(x)=\frac{1}{2\sqrt{x}}+\frac{5}{x^2}+8x^3\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\displaystyle\frac{1}{2\sqrt{x}}+\frac{5}{x^2}+8x^3 \)
\(\textbf{15)}\) \( f(x)=4e^x-3\ln x+x^2 \)
The derivative is \( f'(x)=4e^x-\displaystyle\frac{3}{x}+2x \)
\(\,\,\,\,\,\text{Differentiate each term separately.}\)
\(\,\,\,\,\,\frac{d}{dx}(4e^x)=4e^x\)
\(\,\,\,\,\,\frac{d}{dx}(-3\ln x)=-\frac{3}{x}\)
\(\,\,\,\,\,\frac{d}{dx}(x^2)=2x\)
\(\,\,\,\,\,f'(x)=4e^x-\frac{3}{x}+2x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=4e^x-\displaystyle\frac{3}{x}+2x \)
\(\textbf{16)}\) \( f(x)=6\cos x-2\sin x+5x \)
The derivative is \( f'(x)=-6\sin x-2\cos x+5 \)
\(\,\,\,\,\,\text{Differentiate each trig and linear term separately.}\)
\(\,\,\,\,\,\frac{d}{dx}(6\cos x)=-6\sin x\)
\(\,\,\,\,\,\frac{d}{dx}(-2\sin x)=-2\cos x\)
\(\,\,\,\,\,\frac{d}{dx}(5x)=5\)
\(\,\,\,\,\,f'(x)=-6\sin x-2\cos x+5\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=-6\sin x-2\cos x+5 \)
\(\textbf{17)}\) \( f(x)=\sec^2 x+3\tan x \)
The derivative is \( f'(x)=2\sec^2 x\tan x+3\sec^2 x \)
\(\,\,\,\,\,\text{Differentiate each term separately.}\)
\(\,\,\,\,\,\frac{d}{dx}(\sec^2 x)=2\sec x(\sec x\tan x)\)
\(\,\,\,\,\,\frac{d}{dx}(3\tan x)=3\sec^2 x\)
\(\,\,\,\,\,f'(x)=2\sec^2 x\tan x+3\sec^2 x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=2\sec^2 x\tan x+3\sec^2 x \)
\(\textbf{18)}\) \( f(x)=\ln(x)+\log_2(x)+x^3 \)
The derivative is \( f'(x)=\displaystyle\frac{1}{x}+\frac{1}{x\ln 2}+3x^2 \)
\(\,\,\,\,\,\text{Use the logarithmic derivative rules and power rule.}\)
\(\,\,\,\,\,\frac{d}{dx}(\ln x)=\frac{1}{x}\)
\(\,\,\,\,\,\frac{d}{dx}(\log_2 x)=\frac{1}{x\ln 2}\)
\(\,\,\,\,\,\frac{d}{dx}(x^3)=3x^2\)
\(\,\,\,\,\,f'(x)=\frac{1}{x}+\frac{1}{x\ln 2}+3x^2\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=\displaystyle\frac{1}{x}+\frac{1}{x\ln 2}+3x^2 \)
\(\textbf{19)}\) \( f(x)=2^x+5^x-7x \)
The derivative is \( f'(x)=2^x\ln 2+5^x\ln 5-7 \)
\(\,\,\,\,\,\text{Use } \frac{d}{dx}(a^x)=a^x\ln a \text{ and } \frac{d}{dx}(x)=1.\)
\(\,\,\,\,\,\frac{d}{dx}(2^x)=2^x\ln 2\)
\(\,\,\,\,\,\frac{d}{dx}(5^x)=5^x\ln 5\)
\(\,\,\,\,\,\frac{d}{dx}(-7x)=-7\)
\(\,\,\,\,\,f'(x)=2^x\ln 2+5^x\ln 5-7\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=2^x\ln 2+5^x\ln 5-7 \)
\(\textbf{20)}\) \( f(x)=\displaystyle\frac{4}{\sqrt{x}}-3x^{-2}+\sin x \)
The derivative is \( f'(x)=-\displaystyle\frac{2}{x^{3/2}}+\frac{6}{x^3}+\cos x \)
\(\,\,\,\,\,\text{Rewrite } \frac{4}{\sqrt{x}} \text{ as } 4x^{-1/2}. \)
\(\,\,\,\,\,f(x)=4x^{-1/2}-3x^{-2}+\sin x\)
\(\,\,\,\,\,f'(x)=4\left(-\frac{1}{2}\right)x^{-3/2}-3(-2)x^{-3}+\cos x\)
\(\,\,\,\,\,f'(x)=-2x^{-3/2}+6x^{-3}+\cos x\)
\(\,\,\,\,\,f'(x)=-\frac{2}{x^{3/2}}+\frac{6}{x^3}+\cos x\)
\(\,\,\,\,\,\)The derivative is \( f'(x)=-\displaystyle\frac{2}{x^{3/2}}+\frac{6}{x^3}+\cos x \)
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